Let $(\Omega,\mathcal{F},\mathbb{P}):=(M^{\mathbb{N}_{0}},\mathcal{M}^{\mathbb{N}_{0}},\mathbb{P})$ be a probability space where $M=\left\{0,1,2,3,4\right\}$, $\mathcal{M}^{\mathbb{N}_{0}}$ is product $\sigma$-algebra. Note that if we consider the cylinder $I=\left[x_{0},x_{1},\ldots,x_{k}\right]$ defined by
\begin{equation}I=\left[x_{0},x_{1},\ldots,x_{n}\right]:=\left\{\mathbf{y}=(y_{i})_{i\in\mathbb{N}}\in M^{\mathbb{N}_{0}} \left|y_{i}=x_{i} \mbox{ para }i=0,1,\ldots,n \right. \right\}.\end{equation}
Then the collection of all cylinders form a semi-algebra that we called $\mathcal{S}$, futhermore, by measure theory we know that \begin{equation}\mathcal{M}^{\mathbb{N}_{0}}=\sigma(\mathcal{S}).\end{equation}
In this sense, given a fixed transition matrix $\Pi$ of size $5\times 5$ and your invariant measure $\pi$ we define $\mathbb{P}$ in $S$ as
$$\mathbb{P}(I):=\pi(x_{0})\Pi(x_{0},x_{1})\Pi(x_{1},x_{2})\cdots \Pi(x_{k-1},x_{k}).$$

My attempt: Considering the Theorem 6.15 (Furstenberg-Kesten) in this link we should calculate $\widehat{\Omega}$, I think that $$\widehat{\Omega}:=\left\{\mathbf{y}=(y_{n})_{n\in\mathbb{N}}\in M^{\mathbb{N}} | y_{i}=0 \mbox{ for some }i\right\}.\tag{1}$$
In this sense, we know that the singular values of $A_{n}$ are $$\delta_{1}(A_{n})=2^{n+1} \qquad \mbox{ and }\qquad \delta_{2}(A_{n})=Y_{n}Y_{n-1}\cdots Y_{1}Y_{0}.$$
Then, for any $\omega\in \widehat{\Omega}$ the Lyapunov exponents are
$$\Lambda_{1}(\omega)=\lim_{n\rightarrow \infty} \frac{1}{n} \log\delta_{1}(A_{n})=\log(2) $$
and for $n$ large we have